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Buffalo Bills linebacker A.J. Klein named AFC Defensive Player of the Week

Rewarding a strong week from the Bills veteran.

After pacing the way for the Buffalo Bills’ defense in their win over the Los Angeles Chargers, linebacker A.J. Klein has been named the AFC Defensive Player of the Week. He earned the honor in the wake of a career-high 14 tackles, three tackles for loss, 1.5 sacks, and a pass breakup. Now with five sacks, Klein leads the team in that stat.

It’s the second time this season that a Bills defender has earned the award, after Jerry Hughes in Week 7. That one came from the win over the New York Jets. It’s also the first time a Bills linebacker has won the honor since Matt Milano in week three of 2018—yep, neither Milano nor Tremaine Edmunds have earned it since then.

Surprisingly, Chargers end Joey Bosa was not selected for the honor, despite a much more impressive stat line: nine tackles, six tackles-for-loss, three sacks, a pass defense, and a fumble recovery. Bosa dominated the Bills from start to finish. Maybe the NFL was more inclined to give the award to a player whose team won the game?

Either way, congratulations to Klein, who has strung together a few great performances in the last few weeks.